Fig. 226. Resultant Forces Acting on Voussoirs of a Full-Centered Arch.

We shall assume that the arch carries a railroad track and a heavy class of traffic. The weight of roadbed and track may be computed as follows: The ties are to be 8 feet long; the weight of the roadbed and track (and also the live load) is assumed to be distributed over an area 8 feet wide.

 Two rails at 100 pounds per yard will weigh, per square foot of surface..... 8.4 lbs. Oak ties, weighing 150 pounds per tie, will weigh, per square foot of surface...... 9.4 ,, Weight of ballast at 100 pounds per cubic foot, average depth 9 inches.... 75 ,, Total weight 92.8 lbs.

This is the equivalent of 0.58 foot depth of stone, and we therefore add this uniform depth to the reduced load line for the earth.

A 50-ton freight-car, fully loaded, will weigh 134,000 pounds; with a length between bumpers of 37 feet, this will exert a pressure of about 450 pounds per square foot on a strip 8 feet wide. This is equivalent to 2.8 feet of masonry. We shall therefore consider this as a requirement for uniform loading over the whole arch.

It would be more precise to consider the actual wheel loads for the end trucks of two such cars which are immediately following each other; but since the effect of this would be even less than that of the calculation for a locomotive, which will be given later, and since the deep cushion of earth filling will largely obliterate the effect of concentrated loads, the method of considering the loading as uniformly distributed will be used. We therefore add the uniform ordinate equal to 2.8 feet over the whole arch. We shall call this the first condition of loading.